In This post you will get answer of Q.
In Young’s double slit experiment the two slits are d distance apart. Interference pattern is observed on a screen at a distance D from the slits. A dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light is
Solution:
We have the Young’s double slit experiment given by
From the question, we see that the distance between the slits is equal to $d$ and the distance between the slit and screen is equal to $D$.
Hence for the nth dark fringe, we have $ (2n -1) frac{D lambda}{2d} = frac{d}{2}$
Hence, we get $lambda = frac{d^2}{(2n – 1)D} = frac{d^2}{D} , for , n = 1$
Solution:
We have the Young’s double slit experiment given by
From the question, we see that the distance between the slits is equal to $d$ and the distance between the slit and screen is equal to $D$.
Hence for the nth dark fringe, we have $ (2n -1) frac{D lambda}{2d} = frac{d}{2}$
Hence, we get $lambda = frac{d^2}{(2n – 1)D} = frac{d^2}{D} , for , n = 1$